# Element structure of special linear group of degree two over a field

This article gives specific information, namely, element structure, about a family of groups, namely: special linear group of degree two.
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Let $K$ be a field. Consider the special linear group of degree two $SL(2,K)$. The goal of this article is to describe the element structure of $SL(2,K)$.

## Conjugacy class structure

Note that in characteristic two, in the cases of scalar matrices and Jordan block, the eigenvalue of 1 subcase coincides with the eigenvalue of -1 subcase.

COMBINATORIAL BREAKDOWN TABLE: The table below breaks down a collection into various classes or types and provides information on the counts for each type. For some of the columns, totals provide a sanity check that all elements or classes have been accounted for.
Nature of conjugacy class Eigenvalues Characteristic polynomial Minimal polynomial What set can each conjugacy class be identified with? (rough measure of size of conjugacy class) What can the set of conjuacy classes be identified with (rough measure of number of conjugacy classes) What can the union of conjugacy classes be identified with? Semisimple? Diagonalizable over $K$? Splits in $SL_2$ relative to $GL_2$?
Diagonalizable over $K$ with equal diagonal entries, hence a scalar $\{ 1,1 \}$ or $\{ -1,-1\}$ $(x - a)^2$ where $a \in \{ -1,1 \}$ $x - a$ where $a \in \{ -1,1\}$ one-point set two-point set (characteristic not two)
one-point set (characteristic two)
two-point set (characteristic not two)
one-point set (characteristic two)
Yes Yes No
Not diagonal, has Jordan block of size two (in the case of real fields, these are called parabolic conjugacy classes) $1$ (multiplicity 2) or $-1$ (multiplicity 2) $(x - a)^2$ where $a \in \{ -1,1 \}$ Same as characteristic polynomial  ?  ?  ? No No Splits into $|K^\ast/(K^\ast)^2|$ many pieces.
Diagonalizable over a quadratic extension of $K$ but not over $K$. Must necessarily have no repeated eigenvalues. (in the case of real fields, these are called elliptic conjugacy classes) Pair of conjugate elements of a quadratic extension of $K$ of norm 1 $x^2 - ax + 1$, irreducible Same as characteristic polynomial  ?  ?  ? Yes No If $L$ is the quadratic extension, splits into $|K^\ast/N(L^\ast)|$ many pieces.
Diagonalizable over $K$ with distinct (and hence mutually inverse) diagonal entries (in the case of real fields, these are called hyperbolic conjugacy classes) $\lambda, 1/\lambda$ where $\lambda \in K \setminus \{ 0,1,-1 \}$ $x^2 - (\lambda + 1/\lambda)x + 1$ Same as characteristic polynomial  ? quotient space of $K^* \setminus \{ -1,1 \}$ under the inverse map action  ? Yes Yes No
Total NA NA NA NA  ?  ?  ?  ?  ?

### Identification between GL(2)-conjugacy classes and field elements

The trace is a mapping:

Conjugacy classes in $SL(2,K)$ $\to$ $K$

Further, if we are interested in the conjugacy classes of $SL(2,K)$ relative to $GL(2,K)$ (i.e., the relation of being conjugate in $GL(2,K)$, then this mapping is almost bijective: for each of the elements $2,-2 \in K$, there are two inverse images: the scalar matrix and the Jordan block.

Thus, if we are interested in conjugacy with respect to $GL(2,K)$, we can identify:

• Case of characteristic not two: $GL(2,K)$-conjugacy classes in $SL(2,K)$ $\leftrightarrow$ $K \bigsqcup$ a two-point set
• Case of characteristic two: $GL(2,K)$-conjugacy classes in $SL(2,K)$ $\leftrightarrow$ $K \bigsqcup$ a one-point set

In particular, for the size of $K$ a prime power $q$, we have:

• Case of characteristic not two (i.e., odd $q$): There are $q + 2$ conjugacy classes in $SL(2,q)$ relative to $GL(2,q)$
• Case of characteristic two: There are $q + 1$ conjugacy classes in $SL(2,q)$ relative to $GL(2,q)$

This, however, does not classify the conjugacy classes of $SL(2,q)$, because a single $GL(2,q)$-conjugacy class living in $SL(2,q)$ can split in $SL(2,q)$ into multiple conjugacy classes. As noted above, the splitting depends on the sizes of certain quotient groups of the multiplicative group of the field.

### Splitting summary

Here, we provide a brief summary of the splitting of conjugacy classes from $GL_2$ to $SL_2$. As noted in the broad description, there are two types of conjugacy classes that (may) split. More details are in subsequent sections. The key justification in both cases is the splitting criterion for conjugacy classes in the special linear group:

• Conjugacy classes of the Jordan block form: There are two such $GL_2$-conjugacy classes, with representatives:

$\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$

Each of these conjugacy classes splits into $|K^*/(K^*)^2|$ conjugacy classes with respect to $SL_2$. Moreover, if we let $S$ be a transversal (i.e., a system of coset representatives) for $(K^*)^2$ in $K^*$, then the conjugacy class representatives are given by:

$\begin{pmatrix} 1 & a \\ 0 & 1 \\\end{pmatrix}, a \in S$

and:

$\begin{pmatrix} -1 & a \\ 0 & -1 \\\end{pmatrix}, a \in S$

• Conjugacy classes whose characteristic polynomial is an irreducible quadratic with distinct roots in a separable quadratic extension $L$ of $K$. Each such conjugacy class splits into $|K^*/N(L^*)|$ where $N(L^*)$ denotes the image of $L^*$ under the norm map $L \to K$ for the quadratic extension $L$ of $K$.

Note that, for any separable quadratic extension $L$ of $K$, we have $(K^*)^2 \subseteq N(L^*) \subseteq K^*$, so $K^*/N(L^*)$ can be viewed as a quotient of $K^*/(K^*)^2$, viewed as a group.

Note, further, that when the characteristic of the field is not two, the set of possibilities for the extension $L$ of $K$ is itself given by the set of non-identity elements of $K^*/(K^*)^2$.

Below, we summarize some important cases for the field $K$:

Case for field $K$ What can we say about the multiplicative group modulo squares $K^*/(K^*)^2$? (it is always an elementary abelian 2-group) What does it tell us about the Jordan block conjugacy classes? What are the possibilities for separable quadratic extensions $L$ of $K$? What can we say about $K^*/N(L^*)$ for each of these? What can we conclude about the conjugacy class splitting for these?
algebraically closed field of characteristic two trivial group there is only one such conjugacy class, representative $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$ none -- --
algebraically closed field of characteristic not two trivial group two such conjugacy classes: $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$. No splitting relative to $GL_2$ none -- --
perfect field of characteristic two, such as a finite field of characteristic two trivial group there is only one such conjugacy class, representative $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$ depends on the field. For finite fields, there is a unique possibility. trivial group, because $K^* = (K^*)^2$ none of the conjugacy classes split.
finite field of characteristic not two cyclic group:Z2 total of four conjugacy classes, with representatives $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$, $\begin{pmatrix} 1 & a \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & a \\ 0 & -1 \\\end{pmatrix}$ where $a$ is a non-square unique possibility, corresponding to the non-identity element of $K^*/(K^*)^2$ trivial group none of the conjugacy classes split.
real-closed field (basically, similar to the field of real numbers) cyclic group:Z2 total of four conjugacy classes, with representatives $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$, $\begin{pmatrix} 1 & -1 \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & -1 \\ 0 & -1 \\\end{pmatrix}$ unique possibility, namely $\mathbb{C}$. Corresponds to the non-identity element of $\R^*/(\R^*)^2$ cyclic group:Z2, because the image of the norm map takes only positive real values each such conjugacy class splits into two
field of p-adic numbers $\mathbb{Q}_p$ for some prime $p \ne 2$ (note: this field has characteristic zero, not $p$) Klein four-group total of eight conjugacy classes, four with repeated eigenvalue 1 and four with repeated eigenvalue -1 three possibilities, corresponding to the non-identity elements of $K^\ast/(K^\ast)^2$, which is the Klein four-group cyclic group:Z2 for each of the three cases each such conjugacy class splits into two
field of p-adic numbers $\mathbb{Q}_2$ (note: this field has characteristic zero, not $p$) elementary abelian group:E8 total of sixteen conjugacy classes, eight with repeated eigenvalue 1 and eight with repeated eigenvalue -1 seven possibilities, corresponding to the non-identity elements of $K^\ast/(K^\ast)^2$, which is the elementary abelian group:E8 cyclic group:Z2 for each of the three cases each such conjugacy class splits into two
field of rational numbers elementary abelian group with countable basis, with basis elements corresponding to -1 and all the positive primes countably infinite number of conjugacy classes for repeated eigenvalue 1, countably infinite number of conjugacy classes with repeated eigenvalue -1 one for each non-identity element of $K^*/(K^*)^2$ the precise answer varies depending on the choice of the quadratic extension (this is related to deep questions of algebraic number theory/Galois theory). However, it is always elementary abelian with a countable basis. In all cases, each $GL_2$-conjugacy class splits into infinitely many $SL_2$-conjugacy classes.

## Central elements

The center is a subgroup of order either 1 or 2, depending on whether the characteristic of the field is two or not two. In the case of characteristic not equal to two, the center is the subgroup:

$\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix} \right\}$

In the case of characteristic two, the center is the trivial subgroup.

## Jordan block of size two

Over $GL_2$, there are two conjugacy classes of elements of $SL_2$ of this type in characteristic not two. In characteristic two, both of these may split further:

• The conjugacy class of $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$. In particular, this conjugacy class includes all matrices of the form $\begin{pmatrix} 1 & \lambda \\ 0 & 1 \\\end{pmatrix}$ and also all matrices of the form $\begin{pmatrix} 1 & 0 \\ \lambda & 1 \\\end{pmatrix}$ where $\lambda \in K^\ast$.
• The conjugacy class of $\begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$. In particular, this conjugacy class includes all matrices of the form $\begin{pmatrix} -1 & \lambda \\ 0 & -1 \\\end{pmatrix}$ and also all matrices of the form $\begin{pmatrix} -1 & 0 \\ \lambda & -1 \\\end{pmatrix}$ where $\lambda \in K^\ast$.

In characteristic two, both these conjugacy classes collapse into a single one with respect to $GL_2$ because $1 = -1$.

### Splitting criterion

By the splitting criterion for conjugacy classes in the special linear group, we need to do the following:

• First, find the centralizer of any element in the conjugacy class
• Then determine its image under the determinant map to $K^*$
• Then identify the coset space of that image in $K^*$ with the pices into which the $GL_2$-conjugacy class splits in $SL_2$.

The $GL_2$-centralizer of any element with a Jordan block of size two is the subgroup:

$\{ \begin{pmatrix} a & b \\ 0 & a \\\end{pmatrix} \mid a \in K^\ast, b \in K \} \cong K^\ast \times K$

The image of this under the determinant map is:

$\{ a^2 \mid a \in K^\ast \}$

In other words, it is the subgroup of squares in $K^\ast$, denoted $(K^*)^2$.

The cosets of this subgroup in $K^*$ are given by equivalence classes of elements of $K^*$ up to multiplication by squares. The collectin of cosets can be identified with the quotient group $K^*/(K^*)^2$, also called the multiplicative group modulo squares of $K$. Now, if $S$ is a collection of coset representatives of $(K^*)^2$ in $K^*$, consider the set of matrices:

$\begin{pmatrix} c & 0 \\ 0 & 1 \\\end{pmatrix} \mid c \in S \}$

Conjugating a $GL_2$-conjugacy class representative by each of these gives a collection of $SL_2$-conjugacy class representatives. The upshot is that the $SL_2$-conjugacy class representatives for the $GL_2$-conjugacy class of $\begin{pmatrix}1 & 1 \\ 0 & 1 \\\end{pmatrix}$ are:

$\{ \begin{pmatrix} 1 & c \\ 0 & 1 \\\end{pmatrix} \mid c \in S \}$

In particular, the number of such conjugacy classes is $|K^*/(K^*)^2|$.

In the case of characteristic not two, we have another $GL_2$-conjugacy class, the one with representative $\begin{pmatrix}-1 & 1 \\ 0 & -1 \\\end{pmatrix}$. The representatives for the $GL_2$-conjugacy class of $\begin{pmatrix}-1 & 1 \\ 0 & -1 \\\end{pmatrix}$ are:

$\{ \begin{pmatrix} -1 & c \\ 0 & -1 \\\end{pmatrix} \mid c \in S \}$.

In particular, the number of such conjugacy classes is $|K^*/(K^*)^2|$.

The upshot is that:

• For characteristic not equal to two, the set of conjugacy classes with a size two Jordan block can be identified with two disjoint copies of $K^*/(K^*)^2$.
• For characteristic two, the set of conjugacy classes with a size two Jordan block can be identified with a single copy of $K^*/(K^*)^2$.

## Elements diagonalizable over a separable quadratic extension

These elements have pairs of distinct eigenvalues over a separable quadratic extension $L$ of $K$ but do not have eigenvalues over $K$. As we are working in $SL_2$, the eigenvalues must be inverses of each other, and the minimal polynomial is an irreducible polynomial of the form $x^2 - ax + 1$.

There are three parts to the program of understanding such elements:

• Determine all the possible separable quadratic extensions of $K$.
• Determine, for each extension $L$, what are the $GL_2$-conjugacy classes of elements of $SL_2$ whose minimal polynomial is irreducible over $K$ and splits over $L$. This is equivalent to finding all polynomials of the form $x^2 - ax + 1$ whose splitting field is $L$.
• Determine, for each extension $L$, the number of pieces into which any $GL_2$-conjugacy class of the type identified above splits. As we shall see, the splitting behavior depends only on the field, and we do not need to know the precise original $GL_2$-conjugacy class.

### Determination of separable quadratic extensions: characteristic not two

In case the characteristic of the field is not two, there is a correspondence:

Separable quadratic extensions of $K$ (up to equivalence as extensions) $\leftrightarrow$ Non-identity elements of the quotient group $K^*/(K^*)^2$, i.e., non-square elements up to multiplication by squares

The reason is as follows. Any quadratic extension of $K$ is obtained by adjoining a root of an irreducible quadratic. When the characteristic is not two, the quadratic formula shows that this is equivalent to adjoining the square root of the discriminant, so the extension is of the form $K(\sqrt{D})$ for $D$ a non-square in $K^*$. Further $K(\sqrt{D_1})$ and $K(\sqrt{D_2})$ are equivalent as quadratic extensions iff $D_1/D_2$ is a square in $K^*$. Thus, the separable quadratic extensions of $K$ up to equivalence are identified with non-identity cosets of $(K^*)^2$ in $K^*$.

If the characteristic of the field is two, the situation is different, because the separable quadratic extensions are precisely the ones that are not obtained by adjoining square roots. The general theory needs to be filled in.

### Determination of minimal polynomials for a given separable quadratic extension: characteristic not two

Given a separable quadratic extension $L = K(\sqrt{D})$, we first note that there is at least one irreducible quadratic of the form $x^2 - ax + 1$ that splits over this, and hence there exists at least one element of $SL_2$ whose minimal polynomial has splitting field $L$. First note that $D \ne 1$. Consider the polynomial:

$x^2 - \frac{2(D + 1)}{(D - 1)}x + 1$

The roots of this polynomial are:

$\frac{D+1 \pm 2\sqrt{D}}{D - 1}$

Having settled this, we turn to the question of determing the set:

$\{ a \in K \mid x^2 - ax + 1 \mbox{ is irreducible, splits over } K(\sqrt{D})\}$

Using the quadratic formula, we obtain that this is the set:

$\{ a \in K \mid (a^2 - 4)/D \in (K^*)^2 \}$

From a more global perspective, consider the set map:

$K \setminus \{ -2,2\} \to K^\ast/(K^\ast)^2$

given by:

$a \mapsto a^2 - 4 \mbox{ taken modulo } (K^\ast)^2$

We need to exclude $\{ -2,2 \}$ to make sure that the image is an invertible element.

The inverse image of the identity element of $K^\ast/(K^\ast)^2$ corresponds to the minimal polynomials that split over $K$. The inverse image of any non-identity element of $K^\ast/(K^\ast)^2$ corresponds to the minimal polynomials that split over the field extension $L$ corresponding to the non-identity element of $K^\ast/(K^\ast)^2$ by the correspondence detailed above:

Separable quadratic extensions of $K$ (up to equivalence as extensions) $\leftrightarrow$ Non-identity elements of the quotient group $K^*/(K^*)^2$, i.e., non-square elements up to multiplication by squares

In other words, we have a decomposition of $K \setminus \{ -2,2\}$ in terms of fibers above field extensions.

### Determination of splitting criterion: all characteristics

We need to use the splitting criterion for conjugacy classes in the special linear group. In other words:

• We first need to determine the centralizer of any element whose minimal polynomial splits over a separable quadratic extension $L$ of $K$.
• We then need to determine the image of that centralizer under the determinant map to $K^\ast$.
• We then need to identify the coset space of that image and use it to determine the $SL_2$-conjugacy classes into which the $GL_2$-conjugacy class splits.

The first step is easy. We note that the centralizer of this element can be identified with the image of $L^*$ in $GL(2,K)$ via the left action of $L$ on itself, viewing $L$ as a two-dimensional $K$-vector space.

Second, we need to understand the composite map:

$L^* \to GL(2,K) \to K^*$

It turns out that this composite is simply the norm map $N$ for the quadratic extension $L$ of $K$. The map sends $\alpha \in L$ to the product of $\alpha$ and its conjugate.

Thus, the image of the centralizer under the determinant map is the same as the image of $L^*$ under the algebraic norm $N$. Denote this as $N(L^*)$. Then, the coset space can be identified with the group $K^*/N(L^*)$. In particular, each $GL_2$-conjugacy class in $SL_2$ with minimal polynomial having splitting field $L$ must split into $|K^*/N(L^*)|$ many $SL_2$-conjugacy classes.

Further, for $\alpha \in K^*$, $N(\alpha) = \alpha^2$, so $(K^*)^2 = N(K^*) \subseteq N(L^*)$. Thus, the quotient $K^*/N(L^*)$ is a quotient of the quotient $K^*/(K^*)^2$.